4x^2-21x-18 eching

$$4 { x }^{ 2 } -21x-18$$

Omil

$$\left(x-6\right)\left(4x+3\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 4x^{2}+ax+bx-18 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-21 ab=4\left(-18\right)=-72$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -72-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-72 2,-36 3,-24 4,-18 6,-12 8,-9$$

Har bir juftlik yigʻindisini hisoblang.

$$1-72=-71 2-36=-34 3-24=-21 4-18=-14 6-12=-6 8-9=-1$$

Yechim – -21 yigʻindisini beruvchi juftlik.

$$a=-24 b=3$$

4x^{2}-21x-18 ni \left(4x^{2}-24x\right)+\left(3x-18\right) sifatida qaytadan yozish.

$$\left(4x^{2}-24x\right)+\left(3x-18\right)$$

Birinchi guruhda 4x ni va ikkinchi guruhda 3 ni faktordan chiqaring.

$$4x\left(x-6\right)+3\left(x-6\right)$$

Distributiv funktsiyasidan foydalangan holda x-6 umumiy terminini chiqaring.

$$\left(x-6\right)\left(4x+3\right)$$

Kvadrat tenglamalaridan foydalanish qoidalari

Kvadrat koʻp tenglama bu orqali hisoblanadi: ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), bu yerda x_{1} va x_{2} ax^{2}+bx+c=0 kvadrat tenglamaning yechimlari.

$$4x^{2}-21x-18=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni kvadrat formulasi bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat formula ikki yechmni taqdim qiladi, biri ± qo'shish bo'lganda, va ikkinchisi ayiruv bo'lganda.

$$x=\frac{-\left(-21\right)±\sqrt{\left(-21\right)^{2}-4\times 4\left(-18\right)}}{2\times 4}$$

-21 kvadratini chiqarish.

$$x=\frac{-\left(-21\right)±\sqrt{441-4\times 4\left(-18\right)}}{2\times 4}$$

-4 ni 4 marotabaga ko'paytirish.

$$x=\frac{-\left(-21\right)±\sqrt{441-16\left(-18\right)}}{2\times 4}$$

-16 ni -18 marotabaga ko'paytirish.

$$x=\frac{-\left(-21\right)±\sqrt{441+288}}{2\times 4}$$

441 ni 288 ga qo'shish.

$$x=\frac{-\left(-21\right)±\sqrt{729}}{2\times 4}$$

729 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-21\right)±27}{2\times 4}$$

-21 ning teskarisi 21 ga teng.

$$x=\frac{21±27}{2\times 4}$$

2 ni 4 marotabaga ko'paytirish.

$$x=\frac{21±27}{8}$$

x=\frac{21±27}{8} tenglamasini yeching, bunda ± musbat. 21 ni 27 ga qo'shish.

$$x=\frac{48}{8}$$

48 ni 8 ga bo'lish.

$$x=6$$

x=\frac{21±27}{8} tenglamasini yeching, bunda ± manfiy. 21 dan 27 ni ayirish.

$$x=-\frac{6}{8}$$

\frac{-6}{8} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=-\frac{3}{4}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun 6 ga va x_{2} uchun -\frac{3}{4} ga bo‘ling.

$$4x^{2}-21x-18=4\left(x-6\right)\left(x-\left(-\frac{3}{4}\right)\right)$$

p-\left(-q\right) shaklining barcha amallarigani p+q ga soddalashtiring.

$$4x^{2}-21x-18=4\left(x-6\right)\left(x+\frac{3}{4}\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{3}{4} ni x ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$4x^{2}-21x-18=4\left(x-6\right)\times \frac{4x+3}{4}$$

4 va 4 ichida eng katta umumiy 4 faktorini bekor qiling.

$$4x^{2}-21x-18=\left(x-6\right)\left(4x+3\right)$$

Baholash

$$\left(x-6\right)\left(4x+3\right)$$